Find the intersections of two ellipses given by their foci and eccentricities

Question

I am trying to calculate the intersections between two ellipses:

First ellipse has foci: $(-0.2, -0.2), (0.2, 0.2)$ and eccentricity $0.5$.

Second ellipse has foci: $(-0.2, 0.1), (0.3, -0.1)$ and eccentricity $0.5$.

I must find an algorithm that calculates the intersections for each possible pair of them.

Do I first convert these data to the traditional ellipse equations

$$x^2/a^2 + y^2/b^2 = 1?$$

Answer 1

In the comments you reveal you're interested in intersecting ellipses given by foci and eccentricity.

We can use the general equation which you can get from $$\sqrt{(x-\frac1{10})^2+(y-\frac1{10})^2}+\sqrt{(x+\frac1{10})^2+(y+\frac1{10})^2}=2a$$ which we can square a few times to get $$(100a^2-1)y^2-2xy+(100a^2-1)x^2-100a^4+2a^2=0.$$

Use the formula on wikipedia to get the squared eccentricity

$\frac4{((200a^2-2)\operatorname{signum}(100a^4-2a^2)\operatorname{signum}(4(100a^2-1)^2-4)+2)}$

To get this to fit to squared eccentricity $\frac14$ we get $a=\frac{\sqrt{2}}{5},$ i.e. $$7y^2-2xy+7x^2=\frac{392}{625}.$$

To take the two you mention in the comments:

$F_1=(-0.2, -0.2), F_2=(0.2, 0.2), e=0.5$

$$7x^2-2xy+7y^2-48/25=0$$

$F_1=(-0.2, 0.1), F_2=( 0.3, -0.1), e=0.5$

$$1792y^2+320xy-16y+1456x^2-(728x)/5-10001/25=0$$

In maxima CAS:

solve([7*y^2-2*x*y+7*x^2-48/25,1792*y^2+320*x*y-16*y+1456*x^2-(728*x)/5-10001/25],[x,y]);

(%o7) [[x = 0.1663755195470152,y = -0.4733938423206976],
       [x = -0.3113685179502915,y = 0.3789735733435465],
       [x = -0.4260371179039301,y = 0.2497536252287766],
       [x = 0.5184356735695216,y = 0.1789210980994433]]